# Analysis - What are they doing in this proof ?

1. Jun 17, 2007

### quasar987

Analysis - What are they doing in this proof ?!?

1. The problem statement, all variables and given/known data
The thm is "A metric space M is compact iff it is complete and totally bounded".

We have Bolzano-Weierstrass to work with (compact iff sequentially compact).

The "==>" part is direct.

For "<==", they say,

"Let's show that M is sequentially compact. Let {y_k} be a sequence in M and we can assume without loss of generality that all the y_k are distinct."

This is where it gets weird. They show how to construct a Cauchy subsequence that converge but I have no idea what they're doing. They say,

"Given an integer N, cover M with finitely many 1/N-balls, $B_{1/N}(x_{L_1}), ..., B_{1/N}(x_{L_N})$. An infinite number of the y_k lie in one of these balls. Start with N=1. Write $M=B_{1}(x_{L_1})\cup ...\cup B_{1}(x_{L_N})$, and so we can select a subsequence of {y_k} lying entirely in one of these balls. Repeat for N=2, getting a further subsequence lying in a fixed ball of radius 1/2, and so on. Now choose the "diagonal" subsequence, the first member from the first sequence, the second from the second , ans so on. This sequence is Cauchy and since M is complete, it converge."

Firstly what is up with the L's ??? Anyway, my best shot at understanding what he's doing is that he chooses an integer N. Because M is totally bounded, there are finitely many points $\{x_{1},...,x_{L_N}\}$ such that $M=B_{1/N}(x_{1})\cup ...\cup B_{1/N}(x_{L_N})$. There must be at least one x_j such that $B_{1/N}(x_{j})$ contains infinitely many points. Now, for all n<N, $B_{1/N}(x_{j})\subset B_{1/n}(x_{j})$, so there are also infinitely many points is all those balls. For each n, define a subsequence as just those points in $B_{1/n}(x_{j})$.

But...as indicated in the last bit, the convergeant subsequence we're gonna form will consists of the first point of the sequence "spawned" by n=1, the second point will be the second from the sequence spawned by n=2, etc.... the Nth will be the Nth term of the sequence spawned by n=N. But that's as far as we go. Nothing tells us that there is even a single point in $B_{1/N+1}(x_{j})$, so our algorithm stops there it seems.

I probably misunderstood the method though, but I've investigated other interpretations and nothing makes sense.

Last edited: Jun 17, 2007
2. Jun 17, 2007

### morphism

I don't like the subscripts being used. Did you copy them properly? As it stands, I'm understanding that for each N we can cover M with N balls of radius 1/N, which isn't true. There's some inconsistency too, because what you wrote down for N=1 wouldn't make sense anymore.

3. Jun 17, 2007

### quasar987

The part where I quoted the text is exactly as it appears in my book.

I don't like the L_1 either... and that's why I changed it to just "1" in my "interpretation" of the proof (last 3 paragraphs).

4. Jun 17, 2007

### Hurkyl

Staff Emeritus
You could always work the proof out yourself.

The main idea is simply to construct a descending sequence of balls, each of which contains infinitely many terms of your given sequence.

5. Jun 18, 2007

### quasar987

I'm done trying to demystify what they wrote but at least I know how to prove it. It's essentially the same idea as in the proof in R that every bounded sequence has a convergent subsequence. You begin by splitting the bounded area in two, and take a point in one of the halves that hold infinitely many points. Then split that half in two again and select another point in one of the halves with infinitely many points, etc. The resulting subsequence is Cauchy.